Química Orgânica 2 · Adição e condensação aldólica
Claisen–Schmidt condensation — double condensation to dibenzalacetone
Condições: Acetone plus two benzaldehydes in aqueous base under condensation-favoring conditions. Begin with either drawn hydrogen on the right-hand methyl group and the upper-right benzaldehyde, then use either drawn hydrogen on the remaining methyl group and the remaining benzaldehyde to form dibenzalacetone. The product is drawn as (E,E)-dibenzalacetone, the isomer that forms; all four removed alpha hydrogens are explicit. Counterions are not drawn; no stereochemistry is assigned without bond markers.
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O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 15 passos escritos
Move the electrons
The explicit donor alpha proton is removed in a reversible base-catalyzed encounter.
As setas curvas
- The base removes an alpha hydrogen next to the carbonyl.
- The alpha C–H electrons form a π bond to the carbonyl carbon.
- The carbonyl π electrons move onto oxygen.
Move the electrons
These are two resonance forms of the same enolate, not separate molecules. No proton moves between them.
As setas curvas
- An oxygen lone pair forms the carbonyl π bond.
- The π electrons move onto the alpha carbon as a lone pair.
Move the electrons
Enolate carbon attacks the acceptor carbonyl; the donor carbonyl remains intact.
As setas curvas
- The nucleophile uses a lone pair to form the new single bond.
- The π electrons move off carbon, keeping eight electrons around it.
Move the electrons
The retained conjugate water protonates the acceptor-derived alkoxide and regenerates hydroxide.
As setas curvas
- The base uses a lone pair to accept a proton from the O–H bond.
- The bond to hydrogen breaks, and its electrons stay on the atom losing the proton.
Move the electrons
Hydroxide removes the alpha proton next to the carbonyl, forming an enolate. Hydroxide becomes water.
As setas curvas
- The base removes an alpha hydrogen next to the carbonyl.
- The alpha C–H electrons form a π bond to the carbonyl carbon.
- The carbonyl π electrons move onto oxygen.
Move the electrons
These are two resonance forms of the same enolate, not separate molecules. No proton moves between them.
As setas curvas
- An oxygen lone pair forms the carbonyl π bond.
- The π electrons move onto the alpha carbon as a lone pair.
Move the electrons
The enolate loses hydroxide as the C=C bond forms. This gives a neutral enone, with a C=C bond next to C=O.
As setas curvas
- The alpha-carbon lone pair forms a C=C π bond next to the carbonyl.
- The C–O bond electrons leave with oxygen, releasing hydroxide.
Move the electrons
The regenerated hydroxide now removes either drawn hydrogen on the remaining methyl group. The first enone bond is retained rather than attacked.
As setas curvas
- The base removes an alpha hydrogen next to the carbonyl.
- The alpha C–H electrons form a π bond to the carbonyl carbon.
- The carbonyl π electrons move onto oxygen.
Move the electrons
These are two resonance forms of the same enolate, not separate molecules. No proton moves between them.
As setas curvas
- An oxygen lone pair forms the carbonyl π bond.
- The π electrons move onto the alpha carbon as a lone pair.
Move the electrons
The second donor alpha carbon adds to the remaining benzaldehyde molecule.
As setas curvas
- The nucleophile uses a lone pair to form the new single bond.
- The π electrons move off carbon, keeping eight electrons around it.
Move the electrons
The newly formed alkoxide takes a proton from the water made during the second enolate formation.
As setas curvas
- The base uses a lone pair to accept a proton from the O–H bond.
- The bond to hydrogen breaks, and its electrons stay on the atom losing the proton.
Move the electrons
Hydroxide removes the alpha proton next to the carbonyl, forming an enolate. Hydroxide becomes water.
As setas curvas
- The base removes an alpha hydrogen next to the carbonyl.
- The alpha C–H electrons form a π bond to the carbonyl carbon.
- The carbonyl π electrons move onto oxygen.
Move the electrons
These are two resonance forms of the same enolate, not separate molecules. No proton moves between them.
As setas curvas
- An oxygen lone pair forms the carbonyl π bond.
- The π electrons move onto the alpha carbon as a lone pair.
Move the electrons
The enolate loses hydroxide as the C=C bond forms. This gives a neutral enone, with a C=C bond next to C=O.
As setas curvas
- The alpha-carbon lone pair forms a C=C π bond next to the carbonyl.
- The C–O bond electrons leave with oxygen, releasing hydroxide.
Products
Two new C=C bonds flank the retained ketone. The second aldehyde oxygen becomes the final hydroxide, while the original hydroxide oxygen and the first aldehyde oxygen become the two water byproducts. All removed alpha hydrogens and both discarded aldehyde oxygens are still present.
O que você vai praticar
Maintain two independent alpha sites and track the changing hydroxide oxygen through two complete aldol/dehydration cycles.
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